# Simple problems on relativistic energy and momentum

### We will focus on a few simple problems where we will manipulate the equations for relativistic energy and momentum.

This could be seen as a second-year university-level post.

Einstein had shown that the Lorentz transformations were the correct way to switch between the coordinate systems of different frames of reference [1]. He also taught us that Newton’s laws weren’t at all proper relativistic laws. For instance, Newtonian momentum $\mathbf{p} = m \mathbf{v}$, and energy $E = mv^2 / 2$ were not at all accurate at speeds approaching that of light.

Instead, we have all come to learn that the relativistic momentum is written as

$$\label{eq:relativistic momentum} \mathbf{p} = \frac{m \mathbf{v}}{\sqrt{1 – \dfrac{v^2}{c^2}}}.$$

And that the correct relativistic expression for total energy is

$$\label{eq:relativistic energy} E_{\text{tot}} = \frac{mc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}}.$$

We will solve the following problem set:

1. Prove, for a particle travelling at $c$, that the magnitude of the relativistic energy is given by $E = pc$.
2. Show that the energy-momentum relation for a particle with any mass $m$ travelling at any speed $v$ is correct and do mind it is not the famous $E = mc^2$ we are referring to. Use the correct one, if you please.
3. Given that the mass of a proton is $m_p$, calculate its exact speed when its relativistic  translational kinetic energy (which is the relativistic total energy minus its relativistic mass energy) is four times its relativistic mass energy.

### Problem I

Since $E$ is expressed in terms of $p$, we need to rewrite Eq. $\eqref{eq:relativistic momentum}$ by solving for $m$:

$m = \frac{p \sqrt{1 – \dfrac{v^2}{c^2}}}{v}.$

Note, we do not use the vector quantities, just the magnitudes. We can now proceed to substitute this into Eq. $\eqref{eq:relativistic energy}$:

$E_{\text{tot}} = \frac{\left(\dfrac{p \sqrt{1 – \dfrac{v^2}{c^2}}}{v}\right)c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}.$

This reduces to

\begin{align}
E_{\text{tot}} &= \frac{pc^2 \sqrt{1 – \dfrac{v^2}{c^2}}}{v \sqrt{1 – \dfrac{v^2}{c^2}}}, \\
\therefore E_{\text{tot}} &= \frac{pc^2}{v}. \label{eq:E=pc^2/v}
\end{align}

As we are dealing with a particle travelling at speed $c$, we know $v = c$, rendering Eq. $\eqref{eq:E=pc^2/v}$ to

\begin{align}
E_{\text{tot}} &= \frac{pc^2}{c}, \\
\therefore E_{\text{tot}} &= pc.
\end{align}

### Problem II

The energy-momentum relation is

$E^2_{\text{tot}} = p^2c^2 + m^2c^4.$

Substituting Eqs. $\eqref{eq:relativistic momentum}$ and $\eqref{eq:relativistic energy}$, yields

$\left(\frac{mc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2 = \left(\frac{m \mathbf{v}}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2c^2 + m^2c^4,$

which we can continue to work out as follows:

\begin{align*}\left(\frac{mc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2 – \left(\frac{m \mathbf{v}}{\sqrt{1 – \dfrac{v^2}{c^2}}}\right)^2c^2 – m^2c^4 &= 0, \\
\frac{m^2c^4}{1 – \dfrac{v^2}{c^2}} – \frac{m^2v^2c^2}{1 – \dfrac{v^2}{c^2}} – m^2c^4 &= 0, \\
m^2c^4 – m^2v^2c^2 – m^2c^4 + \frac{m^2v^2c^4}{c^2} &= 0, \\
m^2c^4 – m^2c^4 – m^2v^2c^2 + m^2v^2c^2 &= 0, \\
0 – 0 &= 0.
\end{align*}

Hence, for every value of $m$, $p$, and thus $v$, the relation holds.

### Problem III

The relativistic (total) energy is

$E_{\text{tot}} = E_{\text{trans}} + E_{\text{mass}}.$

If the relativistic translational kinetic energy is four times the relativistic mass energy, then we can write

$E_{\text{trans}} = 4E_{\text{mass}}.$

In our case, this then yields for the relativistic (total) energy:

$E_{\text{tot}} = 4E_{\text{mass}} + E_{\text{mass}} = 5E_{text{mass}}.$

To calculate the proton’s speed, we then write

\begin{align*}
\frac{m_pc^2}{\sqrt{1 – \dfrac{v^2}{c^2}}} &= 5E_{\text{mass}} = 5m_pc^2, \\
\frac{1}{\sqrt{1 – \dfrac{v^2}{c^2}}} &= 5, \\
\sqrt{1-v^2/c^2} &= \frac{1}{5}, \\
1-\frac{v^2}{c^2} &= \frac{1}{25}, \\
\frac{v^2}{c^2} &= \frac{24}{25}, \\
v^2 &= \frac{24c^2}{25}, \\
\therefore v &= \sqrt{\frac{24c}{25}} = \frac{2\sqrt{6}c}{5},
\end{align*}

which is about $0.98c$ rounded to two decimals, which means that the proton zips at about 98% of the speed of light through the fabric of the cosmos.

Image Hydrogen bubble chamber Fermilab: Proton with 300 GeV energy producing 26 charged particles in the 30 inch hydrogen bubble chamber at Fermilab. Source: Wikimedia Commons

[1] Einstein, A. (1905) ‘Zur Elektrodynamik bewegter Körper’, Annalen der Physik, 322(10), pp. 891–921. doi: 10.1002/andp.19053221004.

This is a repost. Slight errors in the parsing of LaTeX in the original article of 24 December 2018 have been corrected.